# Impulse Momentum Method for Rotational

1. Aug 6, 2012

### freshbox

1. The problem statement, all variables and given/known data
I have abit of problem understanding this 2 question, I would appreciate if someone can guide me along.

For diagram 1 (Left)
I interpret the condition as: At 0sec, initial velocity is 0, At 5sec final velocity is unknown.

For diagram 2 (Right)
I interpret the condition as: At 0 sec, initial velocity is 0, At 3sec final velocity is 14.5m/s

Please let me know if I get any of the condition wrong, thank you.

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2. Aug 6, 2012

### voko

I think you understand the conditions correctly.

3. Aug 6, 2012

### freshbox

o_0 hmm strange then how come i cannot get the answer...

4. Aug 6, 2012

### voko

What goes wrong?

5. Aug 6, 2012

### freshbox

For Diagram 1 part A

M1=mv
=0

M2=mv
=(20)(V2)
=20V2

M1+I1-2=M2
981-5TA=20V2
5TA=981-20V2
TA=196.2-4V2

Ans is TA=196.2-4V

4V2 and 4V are both different meaning right?

6. Aug 6, 2012

### voko

Why are you ignoring the tangential force?

7. Aug 6, 2012

### freshbox

You mean the 75N force?

What has it got to do with part A?

Last edited: Aug 6, 2012
8. Aug 6, 2012

### voko

You have already solved part A as far as I can tell.

9. Aug 6, 2012

### freshbox

Question Ans is : TA=196.2-4V

Is there any difference between V2 and V? one is initial one is final.

10. Aug 6, 2012

### voko

The initial velocity is zero per the conditions ("released at t = 0"). So v can only be the final.

11. Aug 6, 2012

### freshbox

How about the subscript? Don't have to write?

12. Aug 6, 2012

### voko

Up to you. You can say "let's denote the velocity at t = 5 s as V" or in any other way. That does not change the underlying physics. You just need to know what denotes what.

13. Aug 6, 2012

### freshbox

Part B

M1=Iω
=0

M2=Iω
=(2.5)(V/0.15)
=16.67V

I1-2=Torque x Time
=[-22.5+Ta(0.15)]5
=34.65-3V

M1+I1-2=M2
34.65-3V=16.67V
34.65=19.67V
V=1.76

V=rω
ω=1.76/0.15
ω=11.74 - Ans Wrong

Last edited: Aug 6, 2012
14. Aug 6, 2012

### voko

Hmm. I have used a somewhat different method, but got the same result. Let's see if somebody else here can point out our mistake.

15. Aug 7, 2012

### Staff: Mentor

OK.

Where did you get the 22.5 from? Redo that calculation.

16. Aug 7, 2012

### freshbox

Hi Doc Al, thanks for helping.

I got my -22.5 from the tangential force Ft=75N I used the 75 times 0.3 = 22.5 Because Tension x Radius = Torque

*Ok i think i got it, it's 300mm, 150mm diameters

17. Aug 7, 2012

### Staff: Mentor

There you go.

18. Aug 7, 2012

### freshbox

M1=Iω
=0

M2=Iω
=(2.5)(V/0.15)
=16.67V

I1-2= Torque x Time
=(TaX0.075-75X0.15)5
=17.325-1.5V

M1+I1-2=M2
17.325-1.5v=16.67v
17.325=18.17v
v=0.95

v=rω
0.95=0.15ω
ω=6.3 wrong again

19. Aug 7, 2012

### Staff: Mentor

Same problem with radius vs diameter. (Sorry I didn't spot it before.)

20. Aug 7, 2012

### freshbox

how come i need to take the radius of the axle and not the wheel?